EXPLANATION OF XENON IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) June 10 , 2015 Xenon is a chemical element with symbol Xe and atomic number 54.However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this correct image of Xenon including the following ground state electron configuration: 1s22s22p63s23p63d104s24p64d105s25px25py25pz2 According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (eV) of xenon (from (E1 to E3 ) are the following: E1 = 12.13 , E2 = 21.21, and E3 = 32.12 . For understanding better such ionization energies see also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS. Moreover in “User Kaliambos” you can see my paper “ Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” published in Ind. J. Th. Phys. (2008). ' ' EXPLANATION OF E1 = 12.13 eV = -E(5px2) + E(px1) Here the E(5px2) represents the binding energy of 5px2, while the E(5px1) represents the binding energy of 5px1.The charges (-48e) of (1s22s22p63s23p63d104s2 4p64d105s2) screen the nuclear charge (+54e) and for a perfect screening we would have ζ = 6. Note that the 5px2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5px2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5px1 consists of one electron, we apply the Bohr formula to write E(5px1) = (-13.6057)ζ2/n2 Therefore E1 = 12.13 eV = -E(5px2) + E(5px1) = - (16.95)ζ + 4.1) / n2 Then, using n = 5 the above equation can be written as (13.6057)ζ2 - (16.95) ζ - 299.15 = 0 However, solving for ζ we get ζ = 5.35 < 6, which cannot exist, because the three pairs of 5p make a complete spherical sub-shell leading to a perfect screening with ζ = 6. Under this condition we expect to determine n > 5 . Under this condition we write E1 = 12.13 eV = -E(5px2) + E(5px1) = - (16.95)6 + 4.1) / n2 Then, solving for n we get n = 5.7 > 5 ' '''Of course the two electrons of opposite spin (5px2) do not provide any mutual repulsion, because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion giving a vibration energy, which seems to be like a simple electric repulsion of the Coulomb law. This situation of a vibration energy due to an electromagnetic interaction indeed occurs, because the peripheral velocity of a spinning electron is faster than the speed of light, which invalidates Einstein’s theory of special relativity. (See my FASTER THAN LIGHT). However under the influence of invalid relativity and in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of opposite spin today many physicists believe incorrectly that it is due to the Coulomb repulsion. Under such fallacious ideas I published my paper of 2008. ' ' '''EXPLANATION OF E2 = 21.21 eV = -E(5py2) + E(py1)' Here the E(5py2) represents the binding energy of 5py2, while the E(5py1) represents the binding energy of 5py1. As in the case of E1 the charges (-48e) of (1s22s22p63s23p63d104s24p64d105s2) screen the nuclear charge (+54e) and for a perfect screening we would have ζ = 6. Note that the 5py2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5py2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5py1 consists of one electron, we apply the Bohr formula to write E(5py1) = (-13.6057)ζ2/n2 Therefore E2 = 21.21 eV = -E(5py2) + E(5py1) = - (16.95)ζ + 4.1) / n2 Since n = 5.7 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 685 = 0 Then solving for ζ we get ζ = 7.75 > 6 Here we see that in the absence of one electron the electrons of 5p break the symmetry and lead to the deformations of electron clouds. Thus ζ = 7.75 > 6. EXPLANATION OF E3 = 32.12 eV = -E(5z2) + E(5z1) Here the E(5pz2) represents the binding energy of 5pz2, while the E(5pz1) represents the binding energy of 5pz1. As in the case of E2 the charges (-48e) of (1s22s22p63s23p63d104s24p64d105s2) screen the nuclear charge (+54e) and for a perfect screening we would have ζ = 6. Note that the 5pz2 consists of one pair (2 electrons of opposite spin). Thus applying my formula of 2008 we write -E(5pz2) = -+ (16.95)ζ - 4.1 / n2 On the other hand, since the 5pz1 consists of one electron, we apply the Bohr formula to write E(5pz1) = (-13.6057)ζ2/n2 Therefore E3 = 32.12 eV = -E(5pz2) + E(5pz1) = - (16.95)ζ + 4.1) / n2 Since n = 5.7 the above equation can be written as (13.6057)ζ2 - (16.95)ζ - 1039.48 = 0 Then solving for ζ we get ζ = 9.39 > 6 Here we see that in the absence of two electrons the electrons of 5p break more the symmetry and lead to the great deformations of electron clouds. Category:Fundamental physics concepts